Three charges $$+Q$$, $$q$$, $$+Q$$ are placed respectively, at distance, $$0$$, $$d/2$$ and $$d$$ from the origin, on the $$x$$-axis. If the net force experienced by $$+Q$$, placed at $$x = 0$$, is zero, then value of $$q$$ is:

We have three point charges placed on the $$x$$-axis.

At $$x = 0$$ is the charge $$+Q$$, at $$x = d/2$$ is the charge $$q$$, and at $$x = d$$ is another charge $$+Q$$. Our task is to adjust the value of $$q$$ so that the net electrostatic force acting on the charge $$+Q$$ at the origin becomes zero.

First, we recall Coulomb’s law. It states that the magnitude of the force between two point charges $$Q_1$$ and $$Q_2$$ separated by a distance $$r$$ is

$$F \;=\; k\;\dfrac{|Q_1Q_2|}{r^2},$$

where $$k$$ is Coulomb’s constant. The direction of the force is along the line joining the charges: like charges repel, unlike charges attract.

Now, we calculate the individual forces acting on the charge $$+Q$$ at the origin.

1. Force due to the charge $$q$$ at $$x=d/2$$
The distance between $$+Q$$ at the origin and $$q$$ at $$x=d/2$$ is $$d/2$$.
Hence, the magnitude of that force is

$$F_{q} \;=\; k \,\dfrac{|Q\,q|}{(d/2)^2}.$$

We simplify the denominator:

$$\bigl(d/2\bigr)^2 = \dfrac{d^2}{4},$$

so

$$F_{q} = k \,\dfrac{|Q\,q|}{d^2/4} = \;4k \,\dfrac{|Q\,q|}{d^2}.$$

2. Force due to the charge $$+Q$$ at $$x=d$$
The distance from the origin to this charge is $$d$$, so the magnitude of the force is

$$F_{Q} \;=\; k \,\dfrac{|Q\,Q|}{d^2} = k \,\dfrac{Q^2}{d^2}.$$

Next, we assign directions. The two charges located at $$x>d$$ lie to the right of the origin.

• The charge at $$x=d$$ is $$+Q$$. Since the charge at the origin is also $$+Q$$, the force between them is repulsive, pushing the origin charge toward the left (negative $$x$$-direction).
• For the charge $$q$$ at $$x=d/2$$, the direction depends on the sign of $$q$$. • If $$q$$ were positive, it would also repel $$+Q$$ at the origin, producing another leftward force. • To balance the leftward force from the charge at $$x=d$$, we require a rightward force. That rightward pull can only arise if $$q$$ is negative, because unlike charges attract. Therefore, $$q$$ must indeed be negative.

Let us therefore write $$q = -\,|q|$$, emphasising its negative sign, but keep $$|q|$$ for its magnitude.

The rightward attractive force (toward the positive $$x$$-direction) on the origin charge is then

$$F_{q} = 4k \,\dfrac{Q\,|q|}{d^2}.$$

The leftward repulsive force (toward the negative $$x$$-direction) is

$$F_{Q} = k \,\dfrac{Q^2}{d^2}.$$

For the net force on the origin charge to be zero, these two forces must be equal in magnitude but opposite in direction. Hence we set

$$F_{q} = F_{Q}.$$

Substituting the expressions we have just obtained,

$$4k \,\dfrac{Q\,|q|}{d^2} \;=\; k \,\dfrac{Q^2}{d^2}.$$

We notice that the factor $$k/d^2$$ appears on both sides, so we can divide both sides by $$k/d^2$$ to simplify:

$$4Q\,|q| \;=\; Q^2.$$

Now we solve for $$|q|$$. Dividing both sides by $$4Q$$ gives

$$|q| \;=\; \dfrac{Q}{4}.$$

But we have already established that $$q$$ must be negative. Therefore,

$$q \;=\; -\,\dfrac{Q}{4}.$$

So the required charge $$q$$ is negative one-quarter of $$Q$$.

Hence, the correct answer is Option C.